Mathematics - 2018-11-11 (page 1 of 3)

Luigi M

00:02

Basically the transition functions for $s\times_{S^1} L$ have as image equivalence classes $[a,z]$ where a in Spin^c(n) and z in S^1, but we quotient out the relation [a,z]=[az,1] which then when passing to the adjoint vector bundle gives me the transition function for the tensor product

Mike Miller

Yeah

And the fact that the scalars are central means we can commute all the things we need to

Luigi M

yeah

do you have time for a follow up question?

(I'm following Spin geometry by Lawson and Michelson), this is Example D.10 in their appendix. assume now that M is spin, so we have a principal Spin bundle P. by taking balanced products P \times_{Z/2} L we can form Spin^c structures for M. The claim is similar, If W is my spinor bundle for P, then the adjoint vector bundle to my Spin^c structure is $W\otimes L$

Mike Miller

I think it should again follow by thinking about transition functions, perhaps now with the additional fact (is it a fact?) that the complex Spinor representation is the complexification of the real Spinor representation

I forget my spin geometry, so I am not sure if that is true

Seems like it should be from your statement though

Luigi M

yeah that's what I though but if i try to write down things I'm not anymore allowed to move the entire $S^1$-factor to the left as before, i.e. $[a,z] \neq [az,1]$ i think, since I'm only allowed to move around $\pm 1$

Mike Miller

I think you should first start with L trivial - the spin^c structure induced by the spin structure - and then show that the rest of the statement follows by twisting spin^c structures as above

Luigi M

00:12

Ok, which should follows from the fact that the canonical Spin^c representation restrict to the canonical spin-one right?

Mike Miller

Yeah

Luigi M

ok I think I got it

thanks a lot

Mike Miller

for sure

Ted Shifrin

Heya MikeM

Mike Miller

hey

Joe Shmo

00:49

hi chat

deleted

Ted Shifrin

Hi deleted Shmo.

Joe Shmo

I am asked to compute $\int_{C_R(0)} \dfrac{dz}{(z^2+4)(z-1)^3}$ in terms of $R$. Just to make sure I'm not doing this wrong. For $R<1$, the integral $=0$. For $1 \le R < 2$, apply the residue theorem, and for $R \ge 2$ I get the integral is $0$ again

for the latter interval, i have a handy dandy exercise that states that $\deg q \ge \deg p + 2 \implies \int_{C_R} \frac{p}{q} = 0$ if $C_R$ contains all the poles

Semiclassical

I'd restrict that to $1<R<2$ and $R>2$, since for $R=1,2$ you'd end up with your contour running through a singularity

Érico Melo Silva

hlo chat

Joe Shmo

ok. there's also $R < 1$, which doesn't run through any singularities, hence the integral is $0$

Ted Shifrin

00:56

Agreed. What I would do for $R>2$ is compute the residue at $\infty$. It's very powerful.

heya @Eric !

Semiclassical

yeah, that's another good approach. (good to compute it both ways imo)

Érico Melo Silva

@Ted about to eat at a michelin starred danish place

Ted Shifrin

@Semiclassic: I'd rather not compute 3 different residues if I can do just one :P

Semiclassical

lol

well, one of those residues is pretty trivial :P

Ted Shifrin

OK, agreed.

Semiclassical

00:59

but yeah, doing the other two involves a small computation

Ted Shifrin

I'm a big believer in the residue at infinity, 'cuz then I get to talk about how you really take residues of meromorphic $1$-forms, not of meromorphic functions.

Semiclassical

ah, true

I mostly like it because it gives me a reason to think about the Riemann sphere :)

Ted Shifrin

Sometimes branched coverings of said ...

Érico Melo Silva

@TedShifrin wut do u mean by this

Semiclassical

yeah

Ted Shifrin

01:02

@ÉricoMeloSilva What u mean?

Érico Melo Silva

why does the residue at infinity let u talk about 1-forms

Ted Shifrin

Oh, I'm claiming that it's the residue of a $1$-form that's actually well-defined.

This becomes relephant when you do the Residue Theorem on compact Riemann surfaces.

You should get there soon in G/H.

Érico Melo Silva

i see i see

gotta bounce

Ted Shifrin

Oh, have a great dinner!

Mike Miller

@TedShifrin Still always read this as an orbit

Ted Shifrin

01:13

LOL ... I can't help your proclivities, Mike.

N. Maneesh

Let $\mathscr S$ be a subbasis for a topology $\mathscr T'$ on $X$. Let $\mathscr S$ be a basis for a topology $\mathscr T$ on $X$. How do I prove $\mathscr T=\mathscr T'$. Let $X\in \mathscr T \implies X=\bigcup_{S\in \mathscr S}S$. Since, each $S\in \mathscr S, S=S$ So, $X\in \mathscr T'$.

For reverse inclusion, Let $X\in \mathscr T' \implies X=\bigcup_{\alpha \in \Lambda} \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}, S_{\alpha\beta}\in \mathscr S$ . If $\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}\neq \emptyset$. Let $x\in \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}\implies \exists x\in S_x \in \mathscr S: S_x\subset\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}$.

Daminark

All this mathscr is gonna give me an aneurysm

Ted Shifrin

Well, Demonark, maybe that's doing you a favor in the long run!

Daminark

:0

Ted Shifrin

I certainly don't like the sentence "Let $X\in\mathscr T$."

N. Maneesh

01:18

$\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}=\bigcup_{x\in \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}}S_x$. Hence $X\in \mathscr T$

Am I correct?

Ted Shifrin

Are you trying to take an arbitrary element of $\mathscr T$?

I encourage sentences with far fewer symbols and more words.

N. Maneesh

Yes. sorry. It should be some $U$

Ted Shifrin

Aha.

I don't think you need arbitrary open sets. Basis elements should suffice for any such proof.

Doesn't that simplify things?

Joe Shmo

what is a subbasis?

N. Maneesh

okay

Joe Shmo

01:21

oh yes.

Ted Shifrin

I didn't know if you were being socratic.

N. Maneesh

Subbase

In topology, a subbase (or subbasis) for a topological space X with topology T is a subcollection B of T that generates T, in the sense that T is the smallest topology containing B. A slightly different definition is used by some authors, and there are other useful equivalent formulations of the definition; these are discussed below. == Definition == Let X be a topological space with topology T. A subbase of T is usually defined as a subcollection B of T satisfying one of the two following equivalent conditions: The subcollection B generates the topology T. This means that T is the small...

@TedShifrin me?

Joe Shmo

me

Ted Shifrin

No, JoeShmo.

Joe Shmo

i was. i meant to write to write that in the adjacent chat on the next window

Rithaniel

01:22

Question: What does something need to have in order to be a geometry?

Ted Shifrin

I literally cannot read stuff like you typed, @N.Maneesh. I need words and as few symbols as possible.

Joe Shmo

@N.Maneesh I know what a sbbasis is

Ted Shifrin

@Rithaniel: I have no idea what your context is. And, despite being a geometer, I may not know how to answer it.

N. Maneesh

@TedShifrin okay. I will type again.

Ted Shifrin

No, just think things through to be as efficient as possible. I really don't like this stuff.

Joe Shmo

01:25

its a funny-ly phrased exercise. i want to answer "by definition"

N. Maneesh

@TedShifrin I think $\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}=\bigcup_{x\in \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}}S_x$. Hence $X\in \mathscr T$ is the important step. right?

Ted Shifrin

I have no idea what you're typing.

Seriously.

You're using symbols that haven't even been defined.

Joe Shmo

i second that. i can't parse all of that either

Rithaniel

Okay, then it's a more complicated question than just a given definition? Because I read someone state that a non-Euclidean geometry is just a geometry where the parallel postulate does not hold, and, best I've been able to tell there are five axioms that Euclidean geometry is built on, and I'm unsure if these other four axioms are just prerequisites for a geometry, or if the person who stated that could be wrong.

N. Maneesh

finite intersection of elements of $\mathscr S$ can be the union of elemnts of $\mathscr S$

this thing I was tryiing to prove in the reverse inclusion.

Ted Shifrin

01:31

@Rithaniel: That is reasonable. But there are other notions (look up Klein's notion of geometry on Wiki). I don't know the context of what you're reading, and I don't live in that world of axiomatics for geometry.

N. Maneesh

using the property of basis $x \in B_1\cap B_2(B_1,B_2\in \mathscr S$). There exist $x\in B_x \subset B_1\cap B_2,$ where $B_x$ is a basis element. So I can write the intersection of $B_1$ and $B_2$ is the union of basis elemnts. @TedShifrin This thing I was trying to prove.

Ted Shifrin

Were the subbasis and the basis you were given related at all? Or is this a totally abstract question?

Rithaniel

It's nothing professional, just a person posting something on a comment thread talking about Lovecraft. What constitutes a "non-Euclidean geometry" came up, which got me wondering.

Ted Shifrin

@Rithaniel: Yeah, to most people that would mean the usual axioms without the parallel postulate. But, honestly, I never think of geometry this way.

To me, a non-Euclidean geometry comes from a constant curvature model with non-zero curvature, so the sphere, projective plane, or hyperbolic plane.

Rithaniel

Can you can also have a geometry with the metric requirement being relaxed? Would that still be a geometry?

N. Maneesh

01:40

@TedShifrin Actually, I wanted to prove $\mathscr S$ is a basis for $\mathscr T$. I got this collection is a subbasis for $\mathscr T$.

Ted Shifrin

Are you doing this in a concrete case, or totally abstractly, @N.Maneesh?

@Rithaniel: There are all sorts of finite projective planes, for example, where there's nothing like what I'm talking about. In some settings, a geometry is just a notion of points and lines in a plane. With certain properties.

N. Maneesh

abstractly

Joe Shmo

can you give the exact problem statement

N. Maneesh

I created an excercise like this Let $\mathscr S$ be a subbasis for a topology $\mathscr T'$ on $X$. Let $\mathscr S$ be a basis for a topology $\mathscr T$ on $X$. prove $\mathscr T=\mathscr T'$.

Ted Shifrin

So why can't I just say that I get a basis by taking finite intersections of subbasis elements?

Wait. YOU created it?

Well, as you created it, it's nonsense.

N. Maneesh

01:42

:(

why?

Ted Shifrin

The subbasis gives you a basis. The two bases have to give you the same topology.

Joe Shmo

i was so puzzled by the question

Ted Shifrin

It's nonsense because you have no hypotheses relating the subbasis and basis, although you used the same letter (which I assume is a mistake).

I'm getting too old to be in this chatroom.

Joe Shmo

youre the spirit animal of this chatroom

Ted Shifrin

I resign.

Rithaniel

01:44

Hey, there needs to be someone to tell us when we're being stupid.

Mike Miller

@Rithaniel I generally like to say that geometry is the study of properties and symmetries of rigid things, while topology is ahout flexible things.

Ted Shifrin

Well, at least you should give Klein's definition :P

Rithaniel

Hmmmm, I like that. Kind of a philosophical approach to it.

I'm loosely familiar with topologies at this point. Maybe I'll try to take some abstract geometry courses next semester.

N. Maneesh

sorry! It was otherway round

Rithaniel

Is that even a thing, though? "Abstract geometry"

Ted Shifrin

01:47

There are not such things, typically, @Rithaniel. Typical colleges/universities may have an "axiomatic" geometry course, typically aimed at future secondary teachers (who don't appreciate it), as well as a differential geometry course. But "abstract geometry"?

I recommend you get Pedoe's book, Geometry: A Comprehensive Course.

Mike Miller

One finds a flavor of geometry they like and explore that

Rithaniel

Oh wow, that's only $20

Let's see if I can find a hardcover, though.

Ted Shifrin

There isn't one.

But it's not going to be horribly axiomatic.

Rithaniel

Really? No hardcover? That's unfortunate.

Ted Shifrin

@MikeM: I was proud to include a chapter on affine and projective geometry in my algebra book, to broaden people's horizons. But hardly anyone every reads/teaches it.

@Rithaniel: I don't believe so. Dover reissued it years ago. I have no idea even who the original publisher was.

N. Maneesh

01:52

I created an excercise like this Let $\mathscr S$ be a basis fa topology $\mathscr T$ and if $\mathscr S$ be a subbasis for a topology $\mathscr T'$. then $\mathscr T=\mathscr T'$.

@TedShifrin

Ted Shifrin

Wait, @N.Maneesh. Are you intentionally using the same letter for the basis and subbasis?

N. Maneesh

yes

Ted Shifrin

We missed this hours ago.

OK. So you start with a subbasis that is already a basis.

N. Maneesh

yes

Ted Shifrin

Geez.

N. Maneesh

01:55

sorry for incorrect preposition earlier.

I wanted to prove both generate the same topology.

Ted Shifrin

So you need to show that any finite intersection of subbasis elements, which is a finite intersection of basis elements, is again a union of basis elements. And we know that's true.

You never need to think about the topologies.

N. Maneesh

using the property of basis $x \in B_1\cap B_2(B_1,B_2\in \mathscr S$). There exist $x\in B_x \subset B_1\cap B_2,$ where $B_x$ is a basis element. So I can write the intersection of $B_1$ and $B_2$ is the union of basis elemnts. @TedShifrin This thing I was trying to prove.

Ted Shifrin

Right, and that's true. What do you mean "trying to prove"?

You take the union over all $x\in B_1\cap B_2$. (In principle, you need a finite intersection, not just two, but same thing.)

N. Maneesh

sorry I copied from previous chat

Mike Miller

@TedShifrin Nice, but not surprising.

Ted Shifrin

01:58

LOL, no.

N. Maneesh

Thank you very much @TedShifrin. We met after a long time :)

Ted Shifrin

I always taught several weeks of projective geometry stuff, figuring classic projective geometry (beginnings of algebraic geometry) would be more fun for most people than weeks of Galois theory.

@N.Maneesh We met? :)

N. Maneesh

@TedShifrin yes, occasionally, 5 months ago, I disturbed everyone in this chat with questions :P. Now, I am busy with my lectures. :)

Ted Shifrin

Sorry, I can't keep track unless people are regulars.

N. Maneesh

It's okay.

Ted Shifrin

02:03

howdy loch

Daminark

"More fun than weeks of Galois theory"
:'(

Ted Shifrin

Demonark: Remember that only epsilon of my students even considered going to graduate school in mathematics. ... And I did several weeks of Galois theory.

Fargle

Did someone say Galois theory?

Daminark

:D

Ted Shifrin

And one of my best students ever went to MIT and did a Ph.D. in algebraic geometry, so it didn't hurt him. (Of course, he didn't stay in mathematics. More from the way the economy works and having a family and kids to support.)

heya @Fargle

Jacksoja

02:07

How can we replace the additive inverse condition of a v.s

Daminark

Yeah, probably gonna pull a lot of us out of academia

Ted Shifrin

Why replace it, @Jacksoja?

Jacksoja

with 0.v =0 for all v inV ?

Fargle

Heya @Ted.

Jacksoja

@TedShifrin it is a question I have,

Ted Shifrin

02:07

Oh ... Then you can do it, @Jacksoja.

What should be $-v$?

Daminark

Projective geometry definitely is nifty

Jacksoja

I mean by replace what do they mean ?

Fargle

I really should go back and try to learn Galois theory. I think I'm strong enough on the other algebra to grasp the basics of it now, and that was the portion of algebra that I was skipping back when my life was less together.

Ted Shifrin

They mean that you should assume $0\cdot v = 0$ and deduce existence of additive inverses.

Jacksoja

is it if we assume one of them we can derive the other ?@TedShifrin

Daminark

02:08

@Fargle d o i t, Galois theory is so much fun

Ted Shifrin

Look at a book by Stewart on Galois theory, @Fargle. It's beautiful.

Or Emil Artin's original pamphlet.

Fargle

Will do. Is it particularly accessible or inaccessible?

Ted Shifrin

@Jacksoja: See above. I already answered that.

Jacksoja

0v= (0+0)v = 0v+0v

Daminark

@Fargle Yes

Jacksoja

02:09

then 0v=0

Ted Shifrin

Not helpful, but try something else.

Knowing $0\cdot v = 0$, what is your candidate for $-v$?

Daminark

(I didn't actually read either, just wanted to take the "Is it X or Y?" "Yes" opportunity)

Jacksoja

-1v ?

Ted Shifrin

Demonark's obnoxiousness quotient doesn't diminish.

Jacksoja

and i need to use the distributive law

Ted Shifrin

02:10

You mean $(-1)v$? @Jacksoja

Jacksoja

Yes

Ted Shifrin

Yes. Do that.

Jacksoja

but it does not seem I proved anything new

Ted Shifrin

New?

Jacksoja

(-1)v+ (1)v = (-1+1)v =0v =0

Ted Shifrin

02:11

You're assuming a different axiom and deducing you have additive inverses.

Right. That's it.

Jacksoja

I mean does not mean I added anything extra to what has been said

Ted Shifrin

They're asking you to show you get the same stuff starting from a different place.

Jacksoja

so in short, 0v=0 forces an additive inverse for all elements in V

Ted Shifrin

Right, whereas you usually use existence of additive inverses to prove that property.

So it's, as you said, an equivalent formulation.

Jacksoja

okay thanks Ted

Ted Shifrin

02:13

Sure.

Jacksoja

so in the other direction

using that we have additive inverse to prove 0v=0 for all v in V

Ted Shifrin

Right

Jacksoja

0v =(0+0)v = 0v+0v , and now we add -0v to both sides

here we had to assume the existence of such thing

-0v right?

Ted Shifrin

Sure.

And then from that you deduce that $(-1)v = -v$.

Jacksoja

these questions i always had trouble with idonno why

Ted Shifrin

02:15

Yeah, this stuff gets really confusing.

Jacksoja

(-1)v =-v is easiar

Ted Shifrin

No it isn't.

Be careful.

Jacksoja

(-1)v+(1)v = (-1+1) v = 0

Ted Shifrin

But you need $0\cdot v = 0$ first.

Jacksoja

hence (-1) v is the additive identity of v

and by uniqness of inverses no ?

Ted Shifrin

02:17

You need first to show $0\cdot v = 0$, then to deduce $(-1)v=-v$.

Jacksoja

okay but i showed that before

Ted Shifrin

But you can't say it's easier.

The first thing is part of it.

Jacksoja

i meant easiar for me not in general

Ted Shifrin

LOL ...

shrugs

Jacksoja

because I understand that property

what it means to be additive inverse

but the rest seems hard, because i usually dont understand what i need to show

Ted Shifrin

02:18

Most of my students tried to use $(-1)v=-v$ to prove $0\cdot v=0$, and that doesn't work.

(From the usual axioms.)

Jacksoja

is there a way to get better at this?

is it important ?

Ted Shifrin

Practice. And this is my least favorite part of math.

Jacksoja

or should I not spend too much time on for now

Ted Shifrin

Right, do real proofs. :)

Jacksoja

okay thanks ! I may return to these once I have time for it

seems something that i know the answer to, but hard to show axiomaticlly

Ted Shifrin

02:21

I am not a big fan of formality in mathematics, but some people are different, and some teachers are different.

Joe Shmo

to what extent not a fan?

Ted Shifrin

A larger extent.

Jacksoja

but the main question is , does it help understanding the subject better?

or is it just a matter of style?

Joe Shmo

for example?

Mike Miller

If that question had a single answer there would be no debate.

Ted Shifrin

02:22

I think these are baby proofs that are far simpler than most complicated proofs that you have to deal with later.

Jacksoja

okay

I take your word on that Ted, if my teacher tells me otherwise , i will give him your number

so await a phone call ^^

Ted Shifrin

LOL, @Jacksoja. Feel free. I don't say that lots of people don't disagree with me. But they're wrong.

Jacksoja

@TedShifrin haha joking, doing self study course,but thanks for the help again

Ted Shifrin

I'm not joking. They're wrong.

MikeM might be on my side. But maybe not.

Mike Miller

I'm mostly on your side. I think it depends on the student. I also think lots of students think that the formality is helping their understanding when in fact it is hiding their lack of understanding.

Ted Shifrin

02:26

Learning formal rules for negating quantifiers makes it "easier," but doesn't help in the long run.

Mike Miller

Mathein is a good example of someone for whom the abstraction helps him understand better, and puts that understanding to good use.

Ted Shifrin

One has to develop intuition.

Mathein is a very sophisticated breed.

And, ultimately, he does have intuitive understanding to communicate most of the time.

Mike Miller

Yes, but I think sometimes built originally from formalities. But note that above I said that most students use it to hide misunderstanding.

Fargle

Like me!

Mike Miller

I think it is rarer for someone to do this effectively.

Ted Shifrin

02:28

We can't all aspire to be Quillens.

Mike Miller

Yeah, Fargle doesn't know anything.

<3

Ted Shifrin

I know Fargle doesn't know anything. He's years behind on diff geo and diff top. :P

Fargle

I don't even know prealgebra don't make fun of me guys I'm only 8

Ted Shifrin

ROFL

Fargle

:)

@MikeMiller <3

Ted Shifrin

02:29

Which brings me to one of my pedagogical concerns. I bring in extra — typically harder or more theoretical — problems to my AoPS kids for the best few to think about. One of my weakest students insists on working on those, gets nowhere, and doesn't work on the more mundane stuff — to his detriment. I've told him twice he needs to do that, but "pride" stops him every time. Sigh.

Should I just not hand it to him? That would be rude. But it might be necessary.

Mike Miller

Give him warning you're thinking of doing that.

Joe Shmo

he has to train the pride out of himself. sooner or later it will dawn on him

Ted Shifrin

We have an exam coming up in two weeks. Originally I was going to include a few of my own more conceptual questions, but now I'm thinking I won't do it.

@JoeShmo: It's tough when you're in 10th or 11th grade.

Joe Shmo

well, you have to hold people up to high standards. and communicate, in the most non-condescending way, that pride will only hold your back

Ted Shifrin

This is a very non-traditional teaching situation. I miss "traditional" college teaching.

Joe Shmo

02:35

if you don't, you're doing him a disservice.

Bob

when I was in college, I remember a conversation I had with a marketing professor

he gave three exams and the students grade for the course was the highest of the three exams

as a result, a lot of students talk more marketing courses and he liked that

Ted Shifrin

I would never grade like that, but that's irrelephant.

Daminark

Hmm, has that kid given much indication as to why he dislikes the mundane problems?

If it's one of those, he's just not interested much by the material in general but once it gets to the tricky stuff he's excited? Desire to not be behind? Etc?

Ted Shifrin

He doesn't look at them, Demonark. It's like he's doing mine to "please" me, even though he gets nowhere, even with my help. And then he struggles on on-line homework. Not good.

Joe Shmo

i guess he marketed his own class

Bob

02:38

his class was very popular

Ted Shifrin

No, I write the problems for the two best students who are more bored with the standard stuff, Demonark.

Bob

the math department was not

Ted Shifrin

I'm not surprised, @Bob. But so what.

Bob

maybe the math department should lower its standards

Joe Shmo

whats the material?

Daminark

02:38

Oh I meant bored in the sense of, maybe he just doesn't even like the subject much but the extra problems are more fun by virtue of being puzzles

But yeah hmm

Ted Shifrin

@Bob, that's destined for failure. The physics and engineering departments will quickly start teaching their own math classes and there won't be any math students. That's just crap. ... Of course, there's no excuse for bad teaching. But incompetent certification on top of it is destined for failure.

Demonark: I don't think he's good enough for that.

Bob

okay

Ted Shifrin

Students who actually need to use the math in their work (e.g., physics and engineering) can't afford to get high grades for incompetence. That's what makes America great. Ideas like that.

On the other hand, I don't excuse bad teaching and bad grading. That does occur too.

Joe Shmo

i was thinking the same thing @Daminark

Mike Miller

i don't know how math departments i've seen can lower their standards any more, given the students i see taking real analysis

Ted Shifrin

02:41

We're talking just about calculus grades here, Mike, but, yeah, you're right. I had students in diff geo who shouldn't have passed Calc I. Seriously.

Joe Shmo

LOL

Ted Shifrin

I'll stick with the students (many not so great) who took many courses from me because they respected the challenge and learned a ton, even if they only got B's and C's.

Daminark

I'm still not sure why the standards got as low as they are. Like, what's stopping universities from just failing a few people out and then everyone else gets the idea and gets their stuff together

Ted Shifrin

I was on university teaching award committees and saw evidence of plenty of teachers like Bob describes. Loved because they made it fun, they were fair, and they made it fun. Nowhere was the word "challenge" or "learning" mentioned.

Demonark: Money. Students are commodities.

Even without MAGA.

Rithaniel

I appreciate when a professor grades work mercilessly. It let's me know what I need to improve upon.

Bob

02:44

I have a friend who was teaching high school math and he was up for tenure

and he was told that he did not use enough colors in his presentations and therefore he was not getting tenure

Ted Shifrin

@Rithaniel: Most professors don't grade the work themselves.

Well, that's total crap, @Bob.

Bob

@TedShifrin That is what he told me and I believe him

Ted Shifrin

Although I do use colors, and I had an ancient colleague, who was to boot a horrid teacher, tell me that boardwork and presentation shouldn't matter in teaching.

@Bob: I'm not not believing him. I'm saying it's crap. And lots of crap occurs at all levels of teaching.

Joe Shmo

@Rithaniel in principle yes, but when you pay to get a degree, do you appreciate it?

Ted Shifrin

@Bob: I'd like to have watched him myself and come to my own conclusions.

Daminark

02:45

I feel like any one student still has more to lose by failing than the uni does by losing them. And I do think that once some failings start, students begin to realize that their future is at risk, they'll start to work

Bob

Well he has a new job today

Rithaniel

Nah, I don't want that at all, @JoeShmo

Ted Shifrin

Demonark: Have you not heard of helicopter parents? (I presume yours are not that.)

Rithaniel

I was actually advocating for the opposite

Ted Shifrin

I'm totally lost now, @Rithaniel.

Daminark

02:47

I have but I feel like they don't really need to be listened to, just hang up if they start to complain

Ted Shifrin

LOL ... then they complain to the dean and to the president. Your view is a bit simplistic.

Rithaniel

I'd appreciate if my professors were to fail me if I kept getting everything wrong.

Bob

speaking of getting things wrong

Rithaniel

If I just pass without learning anything, then what is the purpose of college?

Ted Shifrin

Some of the students I've had who're fondest of me, @Rithaniel, are ones who got bad grades. They never thought of being angry at me for them. But others ....

Daminark

02:48

Why do the dean and president not just ignore them?

Ted Shifrin

@Rithaniel: A credential that says "You may make money now."

Demonark, they can't do that. Money speaks.

Bob

if on Monday I post to Math stack exchange a differential equation of a certain form with an incorrect solution and then somebody points on my error.

and I do the same thing on Wed with a different differential equation of the same form, is that okay?

Ted Shifrin

WTF is your point, Bob?

There's nothing professional going on at MSE, for starters. It's all voluntary, with no professional responsibility or pay.

Bob

I did that and somebody complained

Rithaniel

Yeah, that's a foolish reason to seek a college education, in my view.

Bob

02:50

or at least pointed it out

Ted Shifrin

Well, good that they complained.

But there's no obligation to hold anyone to any standard other than personal integrity.

I complain about lots of incompetence and ultimately just shrug and walk away many times.

But what in the world does this have to do with what we're talking about?

Bob

@TedShifrin thank you for listening

good night

Zee

@TedShifrin am trying to see how the covariant derivative acts on tensors , I write a tensor in its basis representation then apply it , I get the usual big sum of tensor and then the covariant derivative runs through each term just like the lebibniz rule , am I on the right track?

Ted Shifrin

That's too vague for me to answer.

Zee

So I have that the covariant derivative acts on s tensor t by the Leibniz rule

Ted Shifrin

02:55

Yes.

Shouldn't you finally learn some ChatJax to type math?

Zee

Yaaa

I would rather learn math though...

Ted Shifrin

Well, sometimes you have to do both.

Zee

I need to capitalize on the little concentration that I have

Ted Shifrin

And if you're in graduate school, you're going to have to type lots of things — including your thesis work — in LaTeX.

Zee

But your right

Ted Shifrin

02:56

I don't care about your excuses.

Anyhow, yes, you use the Leibniz rule to take the covariant derivative of $s\otimes t$. So?

Zee

I write the tensor in its basis rep

So I get this huge sum of basis vectors and basis covectors with scalers

Ted Shifrin

well, scalar functions

Zee

Ya

Ted Shifrin

so you have to differentiate those scalar functions and the various basis vectors/covectors

So it's a giant product rule.

Zee

Damn , ya

Ted Shifrin

02:59

OK. So what's the question?

Zee

So how does this thing act on covectors?

The covariant derivative

Is it apparent from the def ?

Ted Shifrin

Think about the contraction rule. If $\sigma(s) = \text{constant}$, then $0 = d(\sigma(s)) = \nabla\sigma(s) + \sigma(\nabla s)$.

Zee

And we have functions rather than scalers couse we are dealing with tensor field rather than locally ?

Ted Shifrin

Huh?

No, it's still all local.

Zee

I though when you fix a point on the manifold the functions of the tensor become scalers

Ted Shifrin

03:02

The dual basis has covariant derivative the (transpose of?) the negative of the covariant derivative of the basis.

If we're differentiating, we're working LOCALLY, not AT A POINT.

Zee

In a chart?

Ted Shifrin

Yes.

Zee

Ok I see

Ted Shifrin

I'm going to cook dinner. Night for now, y'all.

Zee

And this thing acting on covectors spits out the negative of whatever it spits out for the dual

?

With a dx instead of a basis vector

Take care

Nick

04:06

$$ \text{distance2banana} = b^3 / \theta^2 $$

Semiclassical

i see you've been to xkcd today

Mog

hello

Kemono Chen

04:46

Which one do you prefer: $\tan^2 x+C$ or $\sec^2 x+C$?

Transcript for

$Mathematics$ Mathematics